What the Y parameters mean. They are the short-circuit admittance parameters, defined by

\(I_1=Y_{11}V_1+Y_{12}V_2, \qquad I_2=Y_{21}V_1+Y_{22}V_2\)

Each one is measured with the other port shorted, which is what makes them quick to read off a picture.

Read the topology. R₁ hangs directly across port 1. The bottom rail is common to both ports, but the top terminal of port 2 is isolated — nothing joins it to the R₁ node. So no current can ever enter or leave the upper terminal of port 2.

Step 1 — the two parameters with V₂ shorted. Set V₂ = 0 and apply V₁. The whole of V₁ sits across R₁:

\(Y_{11}=\left.\dfrac{I_1}{V_1}\right|_{V_2=0}=\dfrac{1}{R_1}\)

Because port 2's upper terminal is open, that current cannot reach it:

\(Y_{21}=\left.\dfrac{I_2}{V_1}\right|_{V_2=0}=0\)

Step 2 — the two parameters with V₁ shorted. Now drive port 2. Since its top terminal connects to nothing, no current flows anywhere:

\(Y_{22}=\left.\dfrac{I_2}{V_2}\right|_{V_1=0}=0, \qquad Y_{12}=\left.\dfrac{I_1}{V_2}\right|_{V_1=0}=0\)

Step 3 — assemble the matrix.

\([Y]=\begin{bmatrix}1/R_1 & 0\\ 0 & 0\end{bmatrix}\)

Where the distractors come from. Option 1 is the answer you would get if R₁ were a shunt element common to both ports, i.e. if the two upper terminals were joined — then every entry becomes 1/R₁ and the matrix is singular. Option 4 is the Y matrix of two independent resistors, one across each port. Option 2 is not even symmetric, so it cannot describe a reciprocal network built only from resistors — a useful one-second check, since reciprocity demands Y₁₂ = Y₂₁.

A note on existence. This network has a perfectly good Y matrix even though its Z matrix does not exist — with port 2 open the impedance parameters run to infinity. That asymmetry is the practical reason both descriptions are kept in circuit theory.

Hence, \([Y]=\begin{bmatrix}1/R_1 & 0\\ 0 & 0\end{bmatrix}\).